

\subsection{\bf The Derivative Method for $\chi^2$---II}

\begin{slide}
\begin{itemize}
\item Now given any guess $\alpha=\alpha_0$, expand $\chi^2$.
\end{itemize}
\[
\chi^2 (\alpha ) = \chi^2 (\alpha_0) + \sum_j (\alpha-\alpha_0)_j\, F_j
+ \sum_{jk} (\alpha -\alpha_0)_j\, (\alpha -\alpha_0)_k\, M_{jk}
\]
\begin{itemize}
\item Adopt a vector notation with $\Delta\alpha_j = \alpha_j -
\alpha_{0j}$.
\begin{equation}
\chi^2 (\Underline{$\alpha$}) = \chi^2 (\Underline{$\alpha$}_0) +
\Delta\Underline{$\alpha$ F} + \Delta\Underline{$\alpha$}^T M
\Delta\Underline{$\alpha$}
\end{equation}

\item Minimum condition is
\begin{eqnarray}
\frac{\partial\chi^2 (\Underline{$\alpha$})}{\partial\alpha_j} & = & 0
\hspace{2em} \mbox{or} \nonumber \\
M\Delta\Underline{$\alpha$} & = & - \Underline{$F$}
\end{eqnarray}

\item So we have new guess
\[
\Underline{$\alpha$} = \Delta\Underline{$\alpha$} + \Underline{$\alpha$}_0 
= \Underline{$\alpha$}_0 - M^{-1} \Underline{$F$}
\]
for true minimum.  

\item Note that as {\bf M} positive definite in Equation~(30), 
this new value of \Underline{$\alpha$} indeed gives a
minimum (not a maximum or saddle point), {\bf as long as second order
Taylor expansion valid}.
\end{itemize}
\end{slide}

